hdoj ztr loves lucky numbers 5676 (dfs模拟)

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ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 918    Accepted Submission(s): 389


Problem Description
ztr loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.

One day ztr came across a positive integer n. Help him to find the least super lucky number which is not less than n.
 

Input
There are T(1n105) cases

For each cases:

The only line contains a positive integer n(1n1018). This number doesn't have leading zeroes.
 

Output
For each cases
Output the answer
 

Sample Input
2450047
 

Sample Output
474747
 

Source
BestCoder Round #82 (div.2)
问题描述
ztr喜欢幸运数字,他对于幸运数字有两个要求1:十进制表示法下只包含4、72:十进制表示法下4和7的数量相等比如47,474477就是而4,744,467则不是现在ztr想知道最小的但不小于n的幸运数字是多少
输入描述
T(1\leq\;T\leq\;10^{5})T(1T105)组数据,每组数据一个正整数nn1\leq\;n\leq\;10^{18}1n1018
输出描述
TT行,每行即答案
输入样例
2450047
输出样例
474747
Hint
请尽可能地优化算法,考虑全面
//没啥说的,就是一个dfs模拟,但是得注意一个特判,见代码
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#define INF 0x3f3f3f3f#define ll long long#define N 100010using namespace std;ll a[N];int cnt;void dfs(int l,int r,int len,ll num){if(l>len/2||r>len/2) return ;if(l+r==len&&l==r){a[cnt++]=num;return ;}if(l<=len/2) dfs(l+1,r,len,num*10+4);if(r<=len/2) dfs(l,r+1,len,num*10+7);}int main(){cnt=0;for(int i=2;i<=18;i+=2)dfs(0,0,i,0);int t;ll n;scanf("%d",&t);while(t--){scanf("%lld",&n);if(n>777777777444444444ll)   //特判 printf("44444444447777777777\n");elseprintf("%lld\n",a[lower_bound(a,a+cnt,n)-a]);}return 0;}

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